# CSE 20 – Discrete Mathematics Dr. Cynthia Bailey Lee Dr. Shachar Lovett Peer Instruction in Discrete Mathematics by Cynthia Leeis licensed under a Creative.

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CSE 20 – Discrete Mathematics Dr. Cynthia Bailey Lee Dr. Shachar Lovett Peer Instruction in Discrete Mathematics by Cynthia Leeis licensed under a Creative Commons Attribution- NonCommercial-ShareAlike 4.0 International License. Based on a work at http://peerinstruction4cs.org. Permissions beyond the scope of this license may be available at http://peerinstruction4cs.org.Cynthia LeeCreative Commons Attribution- NonCommercial-ShareAlike 4.0 International Licensehttp://peerinstruction4cs.org

Today’s Topics: 1. Knights and Knaves 2. Review of Proof by Contradiction 2

1. Knights and Knaves 3

Knights and Knaves  Knights and Knaves scenarios are somewhat fanciful ways of formulating logic problems  Knight: everything a knight says is true  Knave: everything a knave says is false 4

You approach two people, you know the one on the left is a knave, but you don’t know whether the one on the right is a knave or a knight.  Left: “Everything she says is true.”  Right: “Everything I say is true.”  What is she (the one on the right)? A. Knight B. Knave C. Could be either/not enough information D. Cannot be either/situation is contradictory 5

You approach one person, but you don’t know whether he is a knave or a knight.  Mystery person: “Everything I say is true.”  What is he? A. Knight B. Knave C. Could be either/not enough information D. Cannot be either/situation is contradictory 6

You approach one person, but you don’t know whether she is a knave or a knight.  Mystery person: “Everything I say is false.”  What is she? A. Knight B. Knave C. Could be either/not enough information D. Cannot be either/situation is contradictory 7

You meet 3 people: A: “At least one of us is a knave.” B: “At most two of us are knaves.” [C doesn't say anything]  This is a really tricky one, but take a moment to see if you can determine which of the following is a possible solution: A. A: Knave, B: Knave, C: Knave B. A: Knight, B: Knight, C: Knight C. A: Knight, B: Knight, C: Knave (Suggestion: eliminate wrong choices rather than trying to solve the puzzle directly. In your groups: please discuss logic for eliminating choices.) 8

Proof by Contradiction Steps  What are they? A. 1. Assume what you are proving, 2. plug in definitions, 3. do some work, 4. show the opposite of what you are proving (a contradiction). B. 1. Assume the opposite of what you are proving, 2. plug in definitions, 3. do some work, 4. show the opposite of your assumption (a contradiction). C. 1. Assume the opposite of what you are proving, 2. plug in definitions, 3. do some work, 4. show the opposite of some fact you already showed (a contradiction). D. Other/none/more than one. 10

You meet 3 people: A: “At least one of us is a knave.” B: “At most two of us are knaves.” [C doesn't say anything]  A: Knight, B: Knight, C: Knave  Zeroing in on just one of the three parts of the solution, we will prove by contradiction that A is a knight. 11

A: “At least one of us is a knave.” B: “At most two of us are knaves.” [C doesn't say anything] Thm. A is a knight. Proof (by contradiction): Assume not, that is, assume A is a knave. Try it yourself first! 12

A: “At least one of us is a knave.” B: “At most two of us are knaves.” [C doesn't say anything] Thm. A is a knight. Proof (by contradiction): Assume not, that is, assume A is a knave. Then what A says is false. Then it is false that at least one is a knave, meaning zero are knaves. So A is not a knave, but we assumed A was a knave, a contradiction. So the assumption is false and the theorem is true. QED. 13

You meet 3 people: A: “At least one of us is a knave.” B: “At most two of us are knaves.” [C doesn't say anything]  A: Knight, B: Knight, C: Knave  Zeroing in on the second of the three parts of the solution, we will prove by contradiction that B is a knight. 14

A: “At least one of us is a knave.” B: “At most two of us are knaves.” [C doesn't say anything] Thm. B is a knight. Proof (by contradiction): Assume not, that is, assume B is a knave. Try it yourself first! 15

A: “At least one of us is a knave.” B: “At most two of us are knaves.” [C doesn't say anything] Thm. B is a knight. Proof (by contradiction): Assume not, that is, assume B is a knave. Then what B says is false, so it is false that at most two are knaves. So it must be that all three are knaves. Then A is a knave. So what A says is false, and so there are zero knaves. So B must be a knight, but we assumed B was a knave, a contradiction. So the assumption is false and the theorem is true. QED. 16

A: “At least one of us is a knave.” B: “At most two of us are knaves.” [C doesn't say anything] Thm. B is a knight. Proof (by contradiction): Assume not, that is, assume B is a knave. Then what B says is false, so it is false that at most two are knaves. So it must be that all three are knaves. Then A is a knave. So what A says is false, and so there are zero knaves. But all three are knaves and zero are knaves is a contradiction. So B must be a knight, but we assumed B was a knave, a contradiction. So the assumption is false and the theorem is true. QED. 17 We didn’t need this step because we had already reached a contradiction.